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The product of integrable random variables need not be integrable

@Did gave a great example showing that in general the product of two Lebesgue integrable functions need not be integrable. I thought I could just tweak that example and prove this. However, I found the ``tweaking'' is not easy. Suppose random variable $X,Y\in L^{1}$. It means that $E(X),E(Y)<+\infty$ , that is, $\int XdP,\int YdP<+\infty$. I want to find an example where $\int XYdP$ is infinite.

The problem with the definition of expectation is that it is not convenient to calculate. The convenient way of computing the expectation is to use the density function : $E(X)=\int_{-\infty}^{\infty}xf(x)dx$ . However, if I use the density function, then unless $X$ and $Y$ are independent, contructing the density function for $XY$ and making $E(XY)$ infinite will be tricky.

Any suggestion? Thank you.

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As usual: $X=Y$ with density something like $2x^{-3}\cdot[x\gt1]$. (But any integrable independent $X$ and $Y$ yield an integrable product $XY$.) –  Did Mar 4 '12 at 8:37
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For finding the expected value of a function of $X$ and $Y$, you never want to look at the density of that function of $X$ and $Y$. Instead, use the Law of the Unconscious Statistician. –  Robert Israel Mar 4 '12 at 8:58
    
I have been thinking how @Didier Piau came up with good counterexamples. Then, I realized that in both counterexamples he used the properties of $1/x$: $\log(x)$ approaches infinity both at 0 and at infinity. Also, Robert Israel is right: don't mess with joint density function. Instead, since here $X=Y$ by construction, we only need to deal with $X^2$ and only the density function of $X$. Thank you both. –  user16859 Mar 4 '12 at 16:23

1 Answer 1

up vote 3 down vote accepted

Explaining the ingredients of the counterexample by Did:

  1. Taking $X=Y$ does not limit our chances of finding a counterexample, because $XY\le X^2+Y^2$. (So, if $XY$ has infinite expected value, then one of $X^2$ and $Y^2$ does too.)
  2. The focus is on the large values of $X$, because this is where squaring hurts.
  3. The integral that gives the expected value of $X$ should converge, but not too rapidly (like the Gaussian), more like $\int_1^\infty x^{-p}\,dx$ with $p>1$ to be determined.
  4. According to 3, the pdf of $X$ is $c_px^{-p-1}\cdot [x>1]$ where $c_p$ is normalization constant and $[\ \ ]$ the Iverson bracket.
  5. Instead of computing the pdf of $X^2$, we can directly calculate $E(X^2)$ using the pdf of $X$: it is $c_p\int_1^\infty x^2x^{-p-1}\,dx=c_p\int_1^\infty x^{-p+1}\,dx$
  6. The integral in 5 diverges when $p\le 2$. Thus, any $p$ in the range $1< p\le 2$ could be used. We can just as well pick the integer value $p=2$.
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